Unit 4 | unit 4 fluid dynamics

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Introduction to fluid mechanics

Unit - 4

Fluid Dynamics

Q1) Explain surface and body forces and also explain euler equation?

A1)

The motion of the fluid element is influenced by the following forces

Fig no 1 Surface and body forces

1. Normal pressure force

The net pressure force in X direction = p dydz – ( p + p/ X dx) dy dz

= - p/ X dx dy dz

2. Gravity or body force

Let B be the body force per unit mass of fluid having components Bx, By&Bz

In x, y & z directions respectively.

Then, body force acting on the parallelepiped in the direction of X coordinate is = Bxdx dydz

3. Inertia force

The inertia force acting on the fluid mass, along the X coordinates is given by,

Mass x acceleration = dx dydz du/ dy

EQUATION OF MOTION - EULERS EQUATION

Consider steady flow of an ideal fluid along the shown tube.

Separate out a small element of fluid of cross- sectional area d A and length d s from stream tube as a free body from the during fluid.

Fig. shows such a small elements LM of fluid of cross Section area d A and length d s.

Fig no 2 Equation of motion Diagram

Let, p= Pressure of an the elements at L

p + dp = Pressure on the element at M,and

V=Velocity of the fluid element.

The external forces tending to accelerate the fluid elements in the direction of streamline are as follows:

Net pressure force in the direction of flows is, p.dA – (p+dp)dA = -dp. dA

Component of the weight of the fluid elements in the direction of flows is

= - rho.g.dA.ds.cos

=- rho.g.dA.ds (dz/ds) (:. Cos = dz/ds)

= - rho .g.dA.dz

Mass flow of the fluid element = .dA.ds

The Acceleration of the fluid element

a = dv/dt = dv/ds * ds/dt = v. dv/ds

Now , according to Newton’s second law of motion, force= mass x acceleration.

:. – dp. dA.- .g.dA.dz= .dA.ds x v.dv/ds

Dividing both sides by rho .dA, we get

-dp/-. g.dz= v.dv

Or dp/ + v.dv + g.dz=0

This is the required Euler’s equation for motion.

Integrating the Euler’s equation of motion we get

1/ ∫ d p +∫v. d v+ ∫g.dz=constant

p/ ρ + v2 /2 + g z = constant

Dividing by g, we get

p/ g + v 2/2g + z = constant

p /w + v 2/2g + z = constant

Or in other words, P1/ W+V1 2 / 2 g +Z1=P2/ W+V22 / 2g+Z2

Which proves Bernoulli’s equation

Q2) Explain Bernoulli’s equation?

A2)

Assumptions made in Bernoulli’s equation.

The liquid is ideal & incompressible.

The flow is steady & continuous.

The flow is along stream line.

The velocity is uniform over the section & equal to mean velocity.

Only forces acting on the fluid are the gravity forces & pressure forces.

Bernoulli equation

Bernoulli’s equation is derived based on the assumption that fluid is non viscous & therefore frictionless.

Practically, all fluids are real & therefore viscous & there are always some losses in fluid flows.

These losses have, therefore, to be taken into account in the application of Bernoulli’s equation which gets modified as follows

P1/ W+V1 2 / 2 g +Z1=P2/ W+V22 / 2g+Z2 + h L

Where h L= Loss of energy between section 1 & 2.

Applications of Bernoulli’s equation

Bernoulli’s equation is applied is all problems of incompressible fluid flow where energy considerations are involved, But we shall consider its applications to the following measuring devices:

Venturi meter

Pitot- tube

Sub-merged orifice

Orifice meters

Rotameter.

Limitation of Bernoulli’s equation

Velocity of every liquid particle, across any section of pipe is not uniform.

Bernoulli’s equation is not applicable for fluid with unsteady flow.

If fluid flowing in a curved path, the energy due to centrifugal force should be taken into account.

Q3) Explain energy principle?

A3)

Fig no 3 Energy principal determination apparatus

In fluid dynamics, Bernoulli's policy states that fluctuations in fluid flow occur simultaneously with a decrease in stationary pressure or a decrease in the available fluid strength.

The program is named after Daniel Bernoulli who published it in his book Hydrodynamics in 1738. Although Bernoulli found that pressure decreased as the flow rate increased, it was Leonhard Euler who discovered Bernoulli's measurement in a normal way in 1752.

The principle applies only to isentropic flow: where the effects of irreversible processes (such as disorder) and non-adiabatic processes (e.g. heat radiation) are small and ignored.

Bernoulli's system can be applied to a variety of fluid flows, leading to a variety of Bernoulli's calculations; there are different types of Bernoulli's figure for different types of flow.

The simplest form of Bernoulli's equation is valid for unmistakable flow (e.g. high flow of air and gases running through a low March number). Advanced forms can be used for pressurized flow in high March numbers (see Bernoulli statistical availability).

Bernoulli's policy can be based on the principle of energy conservation. This means that, with a constant flow, the sum of all the energies in the liquid along the simplest path is the same at all points in that flow path. This requires that the amount of kinetic energy, potential energy and internal energy remain constant.

Thus an increase in the velocity of a liquid - i.e. an increase in its kinetic energy (strong pressure) - occurs simultaneously a decrease (total energy) of its energy (including constant pressure) and internal energy. When the liquid comes out of a pond, the amount of all kinds of energy is the same in all streamlines because in the pond the energy of each unit (total pressure and gravity ρ g h) is the same everywhere.

Bernoulli's goal can also be found directly in Isaac Newton's Second Movement of Movement. If a small amount of fluid flows horizontally from a high pressure area to a low pressure region, then there is more pressure behind than before. This gives the net power by volume, speeding it up with ease.

Fluid particles are subject to pressure and weight only If the fluid flows horizontally and partially along a certain flow path, where the speed increases it may be due to the fluid in that phase moving from the high-pressure region to the low-pressure region; and if its speed decreases, it may be because it has moved from a low-pressure region to a high-pressure region.

As a result, inside the liquid flowing upwards, the greatest velocity occurs when the pressure is too low, and the lowest velocity occurs when the pressure is too high.

Q4) What is mean by venturimeter?

A4) A venturi meter is a device used for measuring the rate of a flow of a fluid flowing through a pipe it consists of three parts:

A short converging part,

Throat, and

Diverging Part

Fig no 4 Venturimeter

It is based on the principle of Bernoulli’s equation.

Expression for Rate of flow through venturimeter

Consider a venturimeter fitted in a horizontal pipe through which a fluid is following as shown in fig.

Let d 1 = diameter at inlet or at section (1)

P 1= pressure at section (1)

V 1= velocity of fluid at section (1)

A 1= area of section (1) =π/4. D2

And d 2, P 2, V 2, a 2 are corresponding values at section (2)

Applying Bernoulli’s equation at section (1) and (2) we get

P1/ g+V12 /2g+Z1=P2 / g+V22 /2g+Z2

As pipe is horizontal hence z 1 = z2

P1/ g+V12 /2g =P2 / g+V22 /2g

(P1 – p 2 )/ g = V22 /2g - V12 /2g

But (P1 – p 2 )/ g is the difference of pressure heads at section land 2 and it is equal to h

∴(P1 – p 2 )/ g =h

Substituting this value of (P1 – p 2 )/ g in the above eqn we get.

∴ h = V22 /2g - V12 /2g………….. (1)

Now applying continuity equation at section 1 and

a 1 v 1 = a 2 v 2

∴ v 1 = a 2 v 2 / a 1

Substituting this value of v 1 in equation 1

h= V22 /2g – (a 2 v 2 / a 1) 2 / 2 g

h= V22 /2g [ (a 12 –a 22 )/ a 12 ]

V22 = (2 g h a 12 )/ (a 12 –a 22)

Q = a2 v 2

= [a 1 a 2 / ( a 12 –a 22 ) ] * ( 2 g h ) ]

Above equation gives the discharge under ideal conditions and is called, Theoretical discharge Actual discharge will be less than theoretical discharge.

Q act = C d * [a 1 a 2 / ( a 12 –a 22 )] * ( 2 g h ) ]

Where C d = co-efficient of venturimeter and its value is less than 1.

Q5) Explain orifice meter and pitot tube?

A5)

ORIFICE METER

It is a device used for measuring the rate of flow of a fluid through a pipe.

It is a cheaper device compared to venturimeter

It also works on the same principle as that of venturimeters.

It consists of a flat circle plate which has a circular shapes edged holed called orifice, which is concentric with the pipe.

The orifice diameter is kept generally 0.5 times the diameters of the pipe, though it may vary from 0.4 to 0.8 times the pipe diameters.

Fig no 5 Orifice meter

A differential manometers is connect at section

(1) Which is at a distance of about 1.5 to 2.0 times the pipe diameters upstream from the orifice plate, and at section

(2) Which is at a distance of about half the diameters of the orifice on the downstream side from the orifice plate.

PITOT TUBES

It is a device used for measuring the velocity of flow at any point in a pipe or a channel.

It is based on the principle, that if the velocity of flow at a point becomes zero, the pressure there is increased due to the conservation of the kinetic energy into pressure energy.

In its simplest form, the pitot- tube consists of a glass tube, bent at right angles as shown in fig.

I 8.png

Fig no 6 Pitot tubes

The Lower end, which is bent though 90° is directed in the upstream direction as shown in fig.

The liquid rises up in the tube due to the conservation of kinetic energy into pressure energy.

The velocity is determined by measuring the rise of liquid in the tube.

P1= intensity of pressure at point (1)

V1 = velocity of flow at (1)

P2 = pressure at point (2)

V2 = velocity at point (2), which is zero

H = depth of tube in the liquid.

h = rise of liquid in the tube above the fire Surface

Applying Bernoulli’s equation at points (1) and (2)

We get

P1/ W+V1 2 / 2 g +Z1=P2/ W+V22 / 2g+Z2

But Z 1 = Z2as points (1) and (2) are an the same line and V2 =0

P1 /g= pressure Lead at (1) = H

P2 /g = pressure head at (2) = (h +H)

Substituting these values, we get

H+ V1 2 / 2 g = (h+H)

:. h= V1 2 / 2 g or v 1 = (2 g h )

This is theoretical velocity.

Actual velocity is gives by (V 1) = Cv (2 g h)

Where C v= Co-efficient of pitot-tube

:. Velocity at any point V= Cv (2 g h)

Q6) Explain momentum principle?

A6)

It is based on law of conservation of momentum or on the momentum principle, which states that the net force acting on a fluid mass is equal to the change in momentum of flow per unit time in that direction.

The force acting on a fluid mass m is given by the Newtons second law of motion.

F = m x a

Where a is the acceleration acting in the same direction as force F

But a = d v /d t

F = m x d v /d t

= d (m v) /d t, m is constant can be taken inside the differential

Above equation is known as the momentum principle.

It can be written as F d t = d (m v)

Which is known as the momentum equation & states that the impulse of force F acting on a fluid of mass m in a short interval of time dt is equal to the change of momentum d(m v) in the direction of force.

Momentum Correction Factor ()

Momentum correction factor is defined as the ratio of momentum of the flow per second based on actual velocity to the momentum of the flow per second based on average velocity across a section. It is denoted by.

= Momentum per second based on actual velocity/ Momentum per second based on average velocity

= 1 for uniform flow

= 1.01 to 1.07 for turbulent flow in pipes

= 4/3 = 1.33 for laminar flow in pipes.

Q7) Explain forces exerted by fluid flow on pipe bend

A7) The energy produced by the liquid flowing in the pipe - bending

In order to prevent the reflection of the resulting force applied by the liquid flowing into the bend pipe, we will use the basic concept of impulse pressure equation. Before proceeding, it is very important to find and learn the concept of equation intensity.

Let us assume that the liquid flows through a curved pipe as shown here in the following figure. We are looking at two phases here namely phase 1-1 and section 2-2.

Fig no 7 The energy produced by liquid flowing in bending pipe

V1 = Speed of liquid flowing in phase 1

P1 = Pressure fluid pressure in phase 1

A1 = Section 1 location

V2 = Speed of liquid flowing in phase 2

P2 = Pressure fluid pressure in phase 2

A2 = Section 2 location

FX = The energy released by the liquid flowing in the rotation of the pipe in the X-direction.

FY = A police force that causes fluid to flow in a pipe turn in a Y-direction.

As we thought above that Fx and FY are forces, they are produced fluid flowing in the pipe turns X and Y respectively.

In view of Newton's third law of motion, the forces charged by the bending of the pipe in the flowing liquid would be - FX and FY in the direction of X and Y respectively.

There will be more energy and it will work in the flowing liquid. P1A1 and P2A2 are the compressive forces acting on the liquid in phase 1 and phase 2 respectively.

We will now remember the pressure number and will have the next X direction figure.

Net power applied to the liquid in the X-direction = Level of pressure change in the X-direction

P1A1 - P2A2 Cos θ - FX = Mass at each unit time x velocity change

P1A1 - P2A2 Cos θ - FX = ρ Q (Final speed in X direction - First velocity in X-direction)

P1A1 - P2A2 Cos θ - FX = ρ Q (V2 Cos θ - V1)

FX = ρ Q (V1 - V2 Cos +) + P1A1 - P2A2 Cos θ

Similarly, we will remember the pressure figure and will have the following Y direction number.

Net power applies to liquids in Y-direction = Level of pressure change in Y-direction

- P2A2 θ sin - FY = Mass for each unit time x speed change

- P2A2 Sin θ - FY = ρ Q (Final velocity in Y direction - First velocity in Y direction)

- P2A2 θ sin - FY = ρ Q (V2 Sin θ - 0)

FY = ρ Q (-V2 Sin θ) - P2A2 Sin θ

Let us determine the following (FR) forces acting on the bending of the pipe and the bending of the angle with the resulting force (FR) with the horizontal direction.

F = √ ( Fx2 + Fy2 )

Tan θ = Fy/ Fx

Q8) Explain vortex flow in terms of free and forced?

A8)

Fig no 8 Vortex flow

Vortex flow: the circulating weight of a liquid or a liquid flow in a curved path.

Free vortex flow

No external torque or power required. Liquid circulation under certain energies previously given to them. In free vortex machines, the overall energy flow remains constant. There is no power connection between the external source and the flow and any power dissipation of the equipment in motion.

Liquid circulation is due to the preservation of angular pressure.

Velocity equals radius.

For a Free vortex flow

v r = constant or v = c/ r

Free vortex flow - Velocity equation

At the center (r = 0) of the velocity rotation approaches infinitely, that point is called a single point.

Free flow of vortex is irrational; therefore, it is also known as irrotational vortex.

In free vortex flow, Bernoulli's equation can be used.

Examples include a whirlpool in a river, water flowing from a bathtub or sink, flowing into the space of a centrifugal pump and flowing around a circular bend of a pipe.

Forced vortex flow

To maintain the forced flow of the vortex, it required a continuous supply of power or external torque.

All liquid particles circulate with an infinite angular velocity ω as a solid body. Therefore, the flow of a forced vortex is called a solid body rotation.

Tangential velocity is directly proportional to the radius.

v = r ω

ω = Angular speed.

r = Radius of particle fluid from the axis of rotation.

The profile of the vortex flow area is illustrated.

Fig no 9 Forced vortex Verified vortex - Height rate

At the forced vortex force each unit weight increases with increasing radius.

The forced vortex is not unthinkable; instead the rotational flow with a constant vorticity 2ω.

Examples of forced vortex flow rotate a vessel containing a liquid with constant angular velocity, flowing within a centrifugal pump.

Q9) Explain the terms

Reynolds number

Froude number

Mach number

Weber number

Euler number

A9) Reynolds number

Reynolds (Re) number helps to predict flow patterns in various fluid flow conditions. In low Reynolds numbers, the flow is usually dominated by laminar (sheet-like) flow, while at high Reynolds numbers the flow is often chaotic.

Confusion is caused by variations in the speed of the curve and direction, which can sometimes combine or incite opposition to the full flow direction (eddy currents). These eddy currents begin to compress the flow, applying energy to the process, which in turn increases the chances of explosion. Reynolds numbers are immeasurably important for liquid mechanics.

Reynolds number has a wide range of applications, ranging from the liquid flow of the pipe to the air flow over the wing of the aircraft. It is used to predict the transition from laminar to turbulent travel, and is used to measure the same flow conditions but with different sizes, such as between aircraft model in the air tunnel and a full-size version.

Early predictor of turbulence and the ability to calculate the measurement results can be used to help predict liquid performance at a larger scale, such as in air and ground or water movement and thus to have climate-related and climate-related results.

Reynolds number is a measure of the weak energy and viscous energy within a fluid that affects the corresponding internal movements due to the different velocities of the fluid. The region in which this force changes function is known as the boundary membrane, as the surface is enclosed within a pipe.

The same effect is achieved by introducing high-velocity fluid currents into low-velocity fluid, like hot gases emitted by flames in the air. The movement associated with this creates a liquid tension, which is a factor in promoting turbulent flow.

The opposite of this effect is the viscosity of the fluid, which usually prevents turbulence. Reynolds' number measures the relative importance of these two types of energy in the given flow conditions, and is an indication of when the turbulent flow will occur in a given situation.

This ability to predict the onset of flow flows is an important invention of mechanical devices such as plumbing or wing systems, but Reynolds' number is also used to measure dynamic water problems, and is used to determine the dynamic similarity between two different fluids, such as between a model aircraft, and its full version.

Such measurements are not online and the use of Reynolds numbers in both cases allows for enhanced rating features.

With respect to laminar and turbulent flow regimes:

laminar flow occurs with low Reynolds numbers, in which viscous forces are prominent, and are characterized by smooth, consistent movements;

Turbulent flow occurs at high Reynolds numbers and is dominated by weak forces, often producing eddy, turbulence and other flow conflicts.

Reynolds' number is described as

Re = u L /

Where:

is the humidity (SI units: kg / m3)

u is flow speed (m / s)

L different line size (m) (see paragraphs below this article for examples)

μ is the strong viscosity of the fluid (Pa · s or N · s / m2 or kg / (m · s))

is the kinematic viscosity of fluid (m2 / s).

Reynolds' number can be defined in a variety of situations where the fluid is a facial-related movement.

These definitions usually include liquid properties of mass and viscosity, as well as the velocity and length of the element or the size of the element (L in the above figure). This size is a matter for the meeting - for example radius and scope work equally well to define sections or circles, but one is chosen by the meeting. On planes or ships, length or width can be used.

The flow of the pipe, or liquid-moving phase, the internal diameter is widely used today. Other shapes such as rectangular pipes or non-circular objects have the same specified width. For liquefaction fluids such as compressed gases or fluid viscosity fluids such as non-Newtonia fluids, special rules apply. Speed can also be a matter of assembly in some cases, especially for renewable vessels.

In practice, matching Reynolds' number is not enough in itself to ensure consistency. The flow of fluid is usually chaotic, and very small changes in the shape and appearance of the surface of the conjunctiva can lead to very different flow. However, Reynolds numbers are a very important guide and are widely used.

Froude number

Froude’s number, Fr, is an infinite number that describes the various states of open channel flow. Froude's number is a measure of gravity and gravity.

Gravity (number) - moves water down

Inertia (denominator) - shows its willingness to do so.

Fr = v / √ (g D)

Where:

V = Water speed

D = Hydraulic depth (flow area / over width)

g = Gravity

When:

Fr = 1, critical flow

Fr> 1, basic flow (fast flow)

Fr <1, poor flow (slow / silent)

Froude's number is the measure of the flow signals such as waves, sand formation, flow / depth of contact in the cross section or between large rocks.

The denominator represents the slower wave surface on the surface of water compared to the speed of the water, called wave celerity. In the flow of critical flow is similar to the flow of speed. Any facial distortion will remain standing. In a continuous flow the flow is controlled from below and the information is transmitted above. This condition leads to post-water effects. Excellent flow is controlled upwards and disturbances are transmitted downstream.

Wave propagation can be used to illustrate these flow conditions: A stick placed in water will form a V-wave pattern below. If the flow is small waves will appear in front of the rod. If the flow is a critical wave it will have a 45o angle. If the flow is excessive the rising waves will appear and the wave angle will be below 45°.

Mach number

Mach (M or Ma) number is an infinite magnitude in the force of a liquid that represents the rate of flow rate beyond the local speed limit.

M = u / c

Where: M local Mack number,

u is the local velocity flow in relation to boundaries (either internal, such as something flow-focused, or external, as a channel), and

c is the speed of the sound in the middle, in the air which varies with the square root of the thermodynamic temperature.

By definition, in March 1, the local flow speed is equal to the speed of the sound. In March 0.65, 65% is subsonic speed, and, on March 1.35, 35% is faster than supersonic speed.

Pilots who use high-performance aircraft use the aircraft number to indicate the actual air conditioning of the vehicle, but the flow path around the vehicle varies in size by three, depending on the local area number.

The local speed of the noise, which is why Mach's number also depends on the temperature of the surrounding gas. Mach number is used primarily to determine the limitations at which the flow can be treated as an inconsistent flow.

The contents can be gas or liquid. The boundary can move in the middle, or it can be set while the center is flowing, or they can both go, at different speeds: what matters is their relative relative speed.

A boundary can be the boundary of an object centered on an internal object, or a channel such as a microphone, design or air channel that transmits a medium. Since Mach's number is defined as a two-speed ratio, it is an infinite number.

If M <0.2-0.3 and the flow are not too volatile and visually impaired, the reduction results will be smaller and easier to measure the flow rate can be used.

Weber number

Weber (We) is a flawless number for liquid technologies that are often useful in analyzing fluid flow where there is a connection between two different liquids, especially multiphase flow with tightly curved surfaces. Named by Moritz Weber (1871–1951).

It can be considered as a measure of the relative importance of fluid inertia compared to groundwater. The quantity helps to test the small flow of the film and the formation of droplets and bubbles.

Euler number

In the statistics, the Euler numbers sequence En numbers (A122045 sequence in OEIS) described the extension of the Taylor series

1/ cosh t = 2 / et + e-t

Where cosh t is hyperbolic cosine .Euler numbers are associated with a specific number of Euler polynomials, namely:

E n = 2 n E n (1/ 2)

Euler's numbers come from the growing series of Taylor's secretive and hyperbolic secant activities. The latter is a function in the definition. They also appear in combinations, especially when calculating the number of exchange permissions for a set with a number of items.

Q10) Explain Buckingham π theorem?

A10)

When a large number of physical variables are involved Rayleigh’s method of dimensional analysis becomes increasingly laborious and cumbersome.

Buckingham’s method is an improvement over the Rayleigh’s method.

Buckingham’s designated the dimensionless group by the letter Π. It is therefore often called Buckingham Π-Method.

The Buckingham’s Π-Method states as follows:

“If there are n variables in a dimensionally homogeneous equation and if these variables contain fundamental dimensions (such as M, L, T, etc.) then the variables are arranged into (n-m) dimensionless terms.

These dimensionless terms are called Π– terms.”

Mathematically, if any variable X1, depends on independent variables, X2, X3, X4,…… Xn; the functional equation may be written as

-------- 1

Equation 1 can also be written as

-------- 2

It is dimensionally homogeneous equation and contains n variables.

If there are m fundamental dimensions, then according to Buckingham’s Π – theorem, it can be written in terms of number of Π – terms in which number of Π terms is equal to (n-m).

Hence equation 2 becomes as

--------3

Each dimensionless – term is formed by combining m variables out of the total n variables with one of the remaining (n-m) variable.

These m variables which appear repeated in each of – terms are consequently called repeating variables and are chosen from among the variables such that they together involve all the fundamental dimensions and they themselves do not form a dimensionless parameter.

Let in the above case X2, X3 and X4 are the repeating variables if the fundamental dimensions m(M, L, T) = 3, then each term is written as:

-------- 4

Where a1, b1, c1, a2, b2, c2 etc. are the constants, which are determined by considering dimensional homogeneity. These values are substituted in equation 4 and values of are obtained. These values of Π’s are substituted in equation 3.

The final general equation for the phenomenon may then be obtained by expressing anyone of the Π – terms as a function of the other as:

Q11) Water is flowing through a tapered pipe having diameters 300 mm and 150 mm at section 1 and 2 respectively. The discharge through the pipe is 40 litre/sec. Section 1 is 10 m above datum and section 2 is 6 m above datum. Find the intensity of pressure at section 2 if that at section 1 is

A11)

Given

Q = 40 litre/sec=0.04 m²/sec

Z₁ = 10m

Z₂ =6m

d₁ =300mm=0.3 m

d2 = 150 mm-0.15m

P1 = 400 kN/m²

To find: P:

Area

By continuity equation,

By Bernoulli's equation

The intensity of pressure of section 2 is 436.84 kN/m²

Q12) A pipe has a length of 200 m with a slope of 1 in 100. The diameter of pipes changes from 1.0 m at higher end to 0.5 m at lower end. Find the pressure at the lower end if the discharge flowing through pipe is 5.00 m/min and pressure at higher end is 45 kN/m².

A12)

Given:

L = 200 m,

P₁ =45 kN/m²

Slope = 1 in 100. z₁ = 0,

d₂ -0.5 m

Q = 5 m/min =5/60 = 0.0833 m/sec

To find: P₂

Slope of pipe is 1 in 100

(Since it is below z₁

i.e., datum, it is considered negative) = -2m

By continuity equation

By Bernoulli's equation

Multiplying by 9.81

Q13) In a vertical pipe conveying oil of specific gravity 0.8, two pressure gauges has been installed at A and B, where diameters are 16 cm and 8 cm respectively. A is 2 m above B. The pressure gauge readings have shown that pressure at B is greater than at A by 0.981 N/cm². Neglecting all losses calculate flow rate

A13)

Given:

Specific gravity S=0.8

Diameter at A, dA = 16 cm = 0.16 m

Diameter at B, dB = 8 cm=0.08 m ZA = 2 m, ZB=0

PB=PA+0/981 N/cm^2

Pressure difference PB-PA= 9.81KN/m^2

Area

By continuity equation,

By Bernoulli's equation

Discharge

Q14) The top and bottom diameters of a 2m long vertical tapering pipe are 100 mm and 50 mm respectively. Water flows down the pipe at 30 liters por second. Find the pressure difference between the two ends of the pipe.

A14)

Given:

Q = 30 Lps = 0.03 m³/sec,

d1 = 0.05 m. d₂ = 0.1 m, z₁ = 0, z₂ = 2 m

Area

By continuity equation,

By Bernoulli's equation

:. The pressure difference between two ends of pipe is 10.415x 10³ Pa

Q15) Obtain an expression for pressure drop from inlet to throat in a horizontally mounted venturimeter in terms of rate of volume flow (0) and inlet diameter (D) for inlet throat diameter ratio is equal to 2.

A15)

d1= Inlet diameter = D

Inlet diameter /Throat diameter= 2

d2 = Throat diameter = D/2

By continuity equation

By Bernoulli's equation

Q16) A venturimeter has an area ratio 9.1, the large diameter being 30 cm. During the flow, the recorded pressure head in the large section is 7.75 m and that at the throat is 5.5 m. Find the discharge through the meter if C,0.98.

A16)

Given:

Discharge of inlet

d1 = 0.3 m

pressure at inlet

Pressure head at throat

Coefficient of discharge Cd, = 0.98

Area

Pressure head

Discharge

Q17) A venturimeter is to be fitted to a pipe of area 0.01787 m² A maximum discharge of 5 m/min under a pressure of 4.5 m flows through the pipe. Find the diameter at the throat for no negative pressure at the throat.

A17)

Given:

a1 = 0.01767 m². Q = 5 m²/min=0.08333 m/sec

Pressure head = 4.5 m

To find: Diameter of throat i.e. d

Considering the venturimeter to be horizontal

Applying Bernoulli’s equation at section (1) and (2)

But Z1=Z2

And as there is no negative pressure at throat considering

Discharge Q

(Assuming C=0.98)

Q18) A venturimeter is fitted to 25 cm diameter pipe. A discharge 10 m³/s flowing through venturimeter, develops a venturi head of 3.5 m of water. Find the minimum diameter of throat so that there is no negative head developed.

A18)

Given:

D = 25 cm = 0.25 m, Q = 10 m³/s,

To find:

d for no negative head developed.

a₁ = area of inlet =

Cd=1

Squaring both side

Q19) A jet of water 50 mm diameter is discharging under a constant head of 70 metres. Find the force exerted by the jet on a fixed plate. Take coefficient velocity as 0.9.

A19)

Given:

d = 50 mm = 0.05 m; H = 70 m

and Cd= 0.9

We know that cross-section area of the jet,

Velocity of the jet,

Force exerted by the jet on the fixed plate,

Q20) A horizontal jet of water is issuing under an effective head of 25 m calculate the diameter of the jet if the force exerted by the jet on a vertical plate is 2.22 KN. Take Cv =1

A20)

Given:

H=25 m; F= 2.22 kN and Cv=1

Let d = Diameter of the jet in meters

We know that cross-section area of the jet,

Velocity of the jet,

Force exerted by the jet on the fixed plate,

Q21) A jet of water of 100 mm diameter, moving with a velocity of 20 m/s striker a stationary plate Find the force on the plate in the direction of the jet, when

(a) The plate is normal to the jet, and

(b) The angle between the jet and plate is 45°.

A21)

Given:

d = 100 mm= 0.1 ni; V=20 m/s and 8 = 45°.

(a) Force on the plate when it is normal to the jet

We known that cross-section area of the jet,

and force on the plate when it is normal to the jet,

(b) Force on the plate when the angle between the jet and plate is 45%

We also known that force on the plate when the angel between the jet the and plate is 45°,

F = av² = 1x sin²45

= 3.142 sin² 45° = 3.142 x (0.707)² = 1.57 kN

Q22) A submarine is cruising at a depth of 20 m in ocean water (density 1020 kg/m). If the forward speed of the submarine is 10 m/s, what readings would be given by pitot and static pressure probes? Assume static probe is located to register free stream static pressure.

A22)

Given:

Speed V = 10 m/s

:. Velocity of submarine

Dynamic pressure

At 20 m depth, static pressure

Pitot Probe pressure = Static pressure + Dynamic pressure

= 200.124 +51

= 251.124 kPa

Q23) A pitot tube records of 7.85 kN/m² as the stagnation pressure when it is head at the centre of pipe of 250 mm diameter conveying water. The static pressure in the pipe is 40 mm of mercury (gauge-vacuum). Calculate the discharge through the pipe assuming that the mean velocity of flow is 0.8 times the maximum velocity. Take C, = 0.98.

A23)

Given:

Stagnation pressure p₁= 7.85 kN/m²

Pressure at center of pipe = 40 mm of mercury

P2 = -0.04×13.6 x 9.81-5.336 kN/m

Mean velocity = 0.8 x maximum velocity

Pressure head

Maximum velocity

Mean velocity (V)=0.8 x Maximum velocity = 0.8 x 5.033=4.0264 m/sec

Discharge Q

Discharge through the pipe 197.65 Lps.

Q24) A pitot static tube is used to measure velocity of an aeroplane. U-tube differential manometer gives deflection of 100 mm of water. If specific weight of air is 12 N/m and coefficient of pitot tube is 0.98. Determine speed of aeroplane. Neglect compressibility effects.

A24)

Given:

C = 0.98. Ya = 12 N/m².

x = 100 mm of water = 0.1m

To find: Speed of aero plane V = ?

Specific gravity of air

Specific gravity of water

Sw = 1

Speed of aircraft

The speed of aeroplane is 39.28 m/s.

Q25) An Orifice of diameter 12 cm is inserted in a pipe of 24 cm diameter. The end pressure gauge fitted upstream and 2 downstream of the orifice metre gives readings 29.43 N/cm and 14.72 N/cm respectively. Co-efficient of discharge of orifice metre is given by 0.6. Find the discharge of water through pipe.

A25)

Given:

d= 24 cm = 0.24 m, p, 29.43 N/cm² = 294.3 kN/m² 4-12 cm-0.12 m. p₂= 14.72 N/cm² Cv= 0.6

Area A₁

Pressure difference

Discharge, Q

The discharge through the pipe is 120 Lps.

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